Reduction of Multiple Subsystems Stability Steady-State Errors Islamic University of Gaza Faculty of Engineering Electrical Engineering Department CH 5,6,7: Reduction of Multiple Subsystems Stability Steady-State Errors Control Systems Design, Dr. Moayed Almobaied

Reduction of Multiple Systems Chapter 5 Reduction of Multiple Systems <<<4.1>>> ###Continuous Time Signals###

Figure 5.2 Components of a block diagram for a linear, time-invariant system

Figure 5.3 a. Cascaded subsystems; b. equivalent transfer function

Figure 5.5 a. Parallel subsystems; b. equivalent transfer function

Figure 5. 6 a. Feedback control system; b. simplified model; c Figure 5.6 a. Feedback control system; b. simplified model; c. equivalent transfer function

Figure 5.7: Block diagram algebra for summing junctions equivalent forms for moving a block a. to the left past a summing junction; b. to the right past a summing junction

Figure 5.8: Block diagram algebra for pickoff points equivalent forms for moving a block a. to the left past a pickoff point; b. to the right past a pickoff point

Block diagram reduction via familiar forms for Example5.1 Problem: Reduce the block diagram shown in figure to a single transfer function

Block diagram reduction via familiar forms for Example5.1 Cont. Steps in solving Example 5.1: a. collapse summing junctions; b. form equivalent cascaded system in the forward path c. form equivalent parallel system in the feedback path; d. form equivalent feedback system and multiply by cascadedG1(s)

Block diagram reduction by moving blocks Example 5.2 Problem: Reduce the block diagram shown in figure to a single transfer function

Steps in the block diagram reduction for Example 5.2 a) Move G2(s) to the left past of pickoff point to create parallel subsystems, and reduce the feedback system of G3(s) and H3(s) b) Reduce parallel pair of 1/G2(s) and unity, and push G1(s) to the right past summing junction c) Collapse the summing junctions, add the 2 feedback elements, and combine the last 2 cascade blocks d) Reduce the feedback system to the left e) finally, Multiple the 2 cascade blocks and obtain final result.

Second-order feedback control system The closed loop transfer function is Note K is the amplifier gain, As K varies, the poles move through the three ranges of operations OD, CD, and UD 0<K<a2/4 system is over damped K = a2/4 system is critically damped K > a2/4 system is under damped

Finding transient response Example 5.3 Problem: For the system shown, find peak time, percent overshot, and settling time. Solution: The closed loop transfer function is And

Gain design for transient response Example 5.4 Problem: Design the value of gain K, so that the system will respond with a 10% overshot. Solution: The closed loop transfer function is For 10% OS we find We substitute this value in previous equation to find K = 17.9

Signal-flow graph components: a. system; b. signal; c Signal-flow graph components: a. system; b. signal; c. interconnection of systems and signals

Building signal-flow graphs a. cascaded system nodes b. cascaded system signal-flow graph; c. parallel system nodes d. parallel system signal-flow graph; e. feedback system nodes f. feedback system signal-flow graph

Problem: Convert the block diagram to a signal-flow graph. Converting a block diagram to a signal-flow graph Problem: Convert the block diagram to a signal-flow graph.

Converting a block diagram to a signal-flow graph Signal-flow graph development: a. signal nodes; b. signal-flow graph; c. simplified signal-flow graph

Mason’s rule - Definitions Loop gain: The product of branch gains found by traversing a path that starts at a node and ends at the same node, following the direction of the signal flow, without passing through any other node more than once. G2(s)H2(s), G4(s)H2(s), G4(s)G5(s)H3(s), G4(s)G6(s)H3(s) Forward-path gain: The product of gains found by traversing a path from input node to output node in the direction of signal flow. G1(s)G2(s)G3(s)G4(s)G5(s)G7(s), G1(s)G2(s)G3(s)G4(s)G5(s)G7(s) Nontouching loops: loops that do not have any nodes in common. G2(s)H1(s) does not touch G4(s)H2(s), G4(s)G5(s)H3(s), and G4(s)G6(s)H3(s) Nontouching-loop gain: The product of loop gains from nontouching loops taken 2, 3,4, or more at a time. [G2(s)H1(s)][G4(s)H2(s)], [G2(s)H1(s)][G4(s)G5(s)H3(s)], [G2(s)H1(s)][G4(s)G6(s)H3(s)]

Mason’s Rule The Transfer function. C(s)/ R(s), of a system represented by a signal-flow graph is Where K = number of forward paths Tk = the kth forward-path gain = 1 - loop gains + nontouching-loop gains taken 2 at a time - nontouching-loop gains taken 3 at a time + nontouching-loop gains taken 4 at a time - ……. = - loop gain terms in that touch the kth forward path. In other words, is formed by eliminating from those loop gains that touch the kth forward path.

Transfer function via Mason’s rule Problem: Find the transfer function for the signal flow graph Solution: forward path G1(s)G2(s)G3(s)G4(s)G5(s) Loop gains G2(s)H1(s), G4(s)H2(s), G7(s)H4(s), G2(s)G3(s)G4(s)G5(s)G6(s)G7(s)G8(s) Nontouching loops 2 at a time G2(s)H1(s)G4(s)H2(s) G2(s)H1(s)G7(s)H4(s) G4(s)H2(s)G7(s)H4(s) 3 at a time G2(s)H1(s)G4(s)H2(s)G7(s)H4(s) Now = 1-[G2(s)H1(s)+G4(s)H2(s)+G7(s)H4(s)+ G2(s)G3(s)G4(s)G5(s)G6(s)G7(s)G8(s)] + [G2(s)H1(s)G4(s)H2(s) + G2(s)H1(s)G7(s)H4(s) + G4(s)H2(s)G7(s)H4(s)] – [G2(s)H1(s)G4(s)H2(s)G7(s)H4(s)] = 1 - G7(s)H4(s) [G1(s)G2(s)G3(s)G4(s)G5(s)][1-G7(s)H4(s)]

Signal-Flow Graphs of State Equations Problem: draw signal-flow graph for: a. place nodes; b. interconnect state variables and derivatives; c. form dx1/dt ; d. form dx2/dt

(continued) e. form dx3 /dt; f. form output Signal-Flow Graphs of State Equations (continued) e. form dx3 /dt; f. form output

Alternate Representation: Cascade Form

Alternate Representation: Cascade Form

Alternate Representation: Parallel Form

Alternate Representation: Parallel Form Repeated roots

Alternate Representation: controller canonical form G(s) = C(s)/R(s) = (s2 + 7s + 2)/(s3 + 9s2 + 26s + 24) This form is obtained from the phase-variable form simply by ordering the phase variable in reverse order

Alternate Representation: controller canonical form System matrices that contain the coefficients of the characteristic polynomial are called companion matrices to the characteristic polynomial. Phase-variable form result in lower companion matrix Controller canonical form results in upper companion matrix

Alternate Representation: observer canonical form Observer canonical form so named for its use in the design of observers G(s) = C(s)/R(s) = (s2 + 7s + 2)/(s3 + 9s2 + 26s + 24) = (1/s+7/s2 +2/s3 )/(1+9/s+26/s2 +24/s3 ) Cross multiplying (1/s+7/s2 +2/s3 )R(s) = (1+9/s+26/s2 +24/s3 ) C(s) And C(s) = 1/s[R(s)-9C(s)] +1/s2[7R(s)-26C(s)]+1/s3[2R(s)-24C(s)] = 1/s{ [R(s)-9C(s)] + 1/s {[7R(s)-26C(s)]+1/s [2R(s)-24C(s)]}}

Alternate Representation: observer canonical form Note that the observer form has A matrix that is transpose of the controller canonical form, B vector is the transpose of the controller C vector, and C vector is the transpose of the controller B vector. The 2 forms are called duals.

Feedback control system for Example 5.8 Problem Represent the feedback control system shown in state space. Model the forward transfer function in cascade form. Solution first we model the forward transfer function as in (a), Second we add the feedback and input paths as shown in (b) complete system. Write state equations

Feedback control system for Example 5.8

State-space forms for C(s)/R(s) =(s+ 3)/[(s+ 4)(s+ 6)]. Note: y = c(t)

Chapter 6 Stability <<<4.1>>> ###Continuous Time Signals###

Stability Definitions A system is stable if the natural response approaches zero as time approaches infinity A system is unstable if the natural response approaches infinity as time approaches infinity A system is marginally stable if the natural response neither decays nor grows but remains constant or oscillates. A system is stable if every bounded input yields a bounded output A system is unstable if any bounded input yields an unbounded output

Common cause of problems in finding closed-loop poles: a. original system; b. equivalent system Stable systems have closed-loop transfer functions with poles in the left half-plane. Unstable systems have closed-loop transfer functions with at least one pole in the right half-plane, and/or poles of multiplicity greater than one on the imaginary axis Marginally stable systems have closed-loop transfer functions with only imaginary axis poles of multiplicity one and poles in the left half-plane.

Routh-Hurwitz Criterion Using this method we can tell how many closed-loop poles are in the left half-plane, in the right half-plane and on the imaginary axis. The method requires two steps: (1) Generate the data table (Routh table) and (2) Interpret the table to determine the number of poles in LHP and RHP.

Initial layout for Routh table

Completed Routh table

Feedback system for Example 6.1; b. Equivalent closed-loop system

Completed Routh table for Example 6.1 Interpretation of Routh table The number of roots of the polynomial that are in the right half-plane is equal to the number of sign changes in the first column.

R-H: Special case. Zero in the first column Replace the zero with , the value of is then allowed to approach zero from –ive or +ive side. Problem Determine the stability of the closed-loop transfer function

R-H: Special case. Zero in the first column Problem Determine the stability of the closed-loop transfer function Solution Write a polynomial that has the reciprocal roots of the denominator

R-H: Special case. Entire row is zero Problem Determine the number of right-half-plane poles in the closed transfer function Solution: Form an auxiliary polynomial, P(s) using the entries of row above row of zeros as coefficient, then differentiate with respect to s finally use coefficients to replace the rows of zeros and continue the RH procedure.

Root positions to generate even polynomials: A , B, C, or any combination An entire row of zeros will appear when a purely even or odd polynomial is a factor of the original polynomial. Even polynomials have roots that are symmetrical about the origin. If we don’t have a row of zeros, we cannot possibly have roots on the imaginary axis.

Routh table for Example 6.5 Problem Determine the number of poles in the right-half-plane, left-half-plan and on the axis for the closed transfer function

Summary of pole locations for Example 6.5

Feedback control system for Example 6.6 Problem Determine the number of poles in the right-half-plane, left-half-plan and on the axis for system Solution: The closed loop transfer function is

Routh table for Example 6.6 2 poles in RHP, 2 poles in LHP no poles on axis The system is unstable

Feedback control system for Example 6.7 Problem Determine the number of poles in the right-half-plane, left-half-plan and on the axis for system Solution: The closed loop transfer function is

Routh table for Example 6.7 Table for polynomial whose roots are the reciprocal of the original 2 sign changes, 2 poles in RHP system is unstable

Feedback control system for Example 6.8 Problem Determine the number of poles in the right-half-plane, left-half-plan and on the axis for system. Draw conclusions about stability of the closed loop system. Solution: The closed loop transfer function is

Routh table for Example 6.8 There is a row of zeros in the s5, so the s6 row forms an even polynomial. 2 sign changes from the even polynomial so 2 poles in the RHP and because of symmetry about origin 2 will be in the LHP The 2 remaining poles are on the axis

Summary of pole locations for Example 6.5

Feedback control system for Example 6.9 Problem Find the range of gain K for the system that will cuase the system to be stable, unstabel, and marginally stable. Assume K>0. Solution: The closed loop transfer function is

Routh table for Example 6.9 For K < 1386 the system is stable. For K > 1386 the system is unstable. For K = 1386 we will have entire row of zeros (s row). We form the even polynomial and differentiate and continue, no sign changes from the even polynomial so the 2 roots are on the axis and the system is marginally stable

Routh table for Example 6.10 Problem Factor the polynomial Solution: from the Routh table we see that the s1 row is a row of zeros. So the even polynomial at the s2 row is since this polynomial is a factor of the original, dividing yields As the other factor so

Stability is State Space Example 6.11 Problem Given the system Find out how many poles in the LHP, RHP and on the axis Solution: First form (sI-A) Now find the det (sI-A) =

Routh table for Example 6.11 One sign change, so 1 pole in the LHP and the system is unstable

Chapter 7 Steady State Error <<<4.1>>> ###Continuous Time Signals###

Test waveforms for evaluating steady-state errors of position control systems Steady-state error is the difference between the input and the output for a prescribed test input as t →∞

Test inputs for steady-state error analysis and design vary with target type

Steady-state error: a. step input; b. ramp input

Closed-loop control system error: a. general representation; b Closed-loop control system error: a. general representation; b. representation for unity feedback systems

System with: a. finite steady-state error for a step input; b System with: a. finite steady-state error for a step input; b. zero steady-state error for step input For pure gain K as in (a), Css= K ess or ess =1/K Css there will always be error But if the pure gain is replaced by an integrator as in (b) the steady state error will be zero.

Feedback control system for Example 7.1 Problem Find the steady-state error for the system in(a) if T(s) = 5/(s2+7s+10) and the input is a unit step. Solution: Using E(s) = R(s)[1-T(s)] We can apply the final value theorem To yield e(∞) =1/2

Steady-state error in terms of G(s) E(s) =R(s) – C(s), C(s) =E(s)G(s) so E(s) = R(s)/1+G(s) Using final value theorem

Steady-state error in terms of G(s) For step input, the steady state error will be zero if there is at least one pure integration in the forward path. For ramp input, the steady state error will be zero if there is at least 2 pure integration in the forward path. For parabolic input, the steady state error will be zero if there is at least 3 pure integration in the forward path.

Steady state error for systems with no integrations Problem Find the steady-state error for inputs 5u(t), 5tu(t), and 5t2u(t) to the system. u(t) is a step function. Solution First verify the closed loop system is stable. For input 5u(t) For input 5tu(t) For input 5t2u(t)

Steady state error for systems with one integration Problem Find the steady-state error for inputs 5u(t), 5tu(t), and 5t2u(t) to the system. u(t) is a step function. Solution First verify the closed loop system is stable. For input 5u(t) For input 5tu(t) For input 5t2u(t)

Static error constants Position constant step input Velocity constant ramp input Acceleration constant parabolic input

Steady state error via static error constants Problem For the system of figure, evaluate the static error constants and find the expected error for standard step, ramp, and parabolic inputs.. Solution First verify the closed loop system is stable. Thus for step input, For ramp input For parabolic input

Steady state error via static error constants Problem For the system of figure, evaluate the static error constants and find the expected error for standard step, ramp, and parabolic inputs. Solution First verify the closed loop system is stable. Thus for step input, For ramp input For parabolic input

Steady state error via static error constants Problem For the system of figure, evaluate the static error constants and find the expected error for standard step, ramp, and parabolic inputs. Solution First verify the closed loop system is stable. Thus for step input, For ramp input For parabolic input

Feedback control system for defining system type System type is the value of n in the denominator, or the number of pure integrations in the forward path.

Relationships between input, system type, static error constants, and steady-state errors

Feedback control system for Example 7.6 Problem Find the value of K so that there is 10% error in the steady state.. Solution: Since the system is Type 1, the error must apply to a ramp input, thus and and K = 672 Applying Routh-Hurwitz we see that the system is stable at this gain.

Feedback control system showing disturbance Applying final value theorem we obtain

Previous system rearranged to show disturbance as input and error as output, with R(s) = 0 Assume step disturbance D(s) = 1/s We get This shows that the steady-state error produced by a step disturbance can be reduced by increasing the dc gain of G1(s) or decreasing the dc gain of G2(s)

Feedback control system for Example 7.7 Problem Find the steady state error due to a step disturbance for the shown system.. Solution: The system is stable, we find

G(s) = G1(s)G2(s) H(s) = H1(s)/G1(s) Forming an equivalent unity feedback system from a general non-unity feedback system G(s) = G1(s)G2(s) H(s) = H1(s)/G1(s)

Non-unity feedback control system for Example 7.8 Problem Find the system type, the appropriate error constant, and the steady state error for a unit step input.. Solution: The system is stable, we convert the system into an equivalent unity feedback system The system is Type 0, and Kp= -5/4. The steady-state error is

Steady-State error for systems in State Space The Laplace transform of the error is E(s) = R(s) –Y(s), and since Y(s) = R(s)T(s). Then E(s) = R(s)[1-T(s)] Using T(s) =Y(s)/U(s) = C(sI-A)-1B + D We have E(s) = R(s)[1-C(sI-A)-1B] Applying final value theorem, we have

Steady-State error for systems in State Space Example: Evaluate the ss error for the system described by Solution: Substitute A,B, and C in For unit step, R(s)=1/s and e(∞)=4/5, for unit ramp, R(s) = 1/s2 and e(∞)= ∞

End of Lecture 5 H.W : 20/2/2018 (with the previous HW) Ch4:8,25,37 Ch5:4,9,27 Ch6:1,19,30 Ch7:13,16,25